What is the degree measure of the least positive angle $x$ for which $\log_2 (\cos x) = −\frac{1}{2}$.
So i rewrote it as:
$\dfrac{\log (\cos x)}{\log 2}=\dfrac{-1}{2}$
but it doesn't seem to open any doors for me. any ideas?
What is the degree measure of the least positive angle $x$ for which $\log_2 (\cos x) = −\frac{1}{2}$.
So i rewrote it as:
$\dfrac{\log (\cos x)}{\log 2}=\dfrac{-1}{2}$
but it doesn't seem to open any doors for me. any ideas?
$ -\frac{1}{2}\log{2} = \log{(1/\sqrt{2})} $. Therefore $$ \log{\cos{x}} = \log{\left(\frac{1}{\sqrt{2}}\right)}. $$ Is it clear now?
Taking $\log x$ to be the natural log,
$$\log (\cos x)=\dfrac{-1}{2}\log 2 $$
$$\log (\cos x)=\log \frac{1}{\sqrt{2}}$$
$$e^{\log (\cos x)}= e^{\log \frac{1}{\sqrt{2}}}$$
Do you see it from here?
Hint: The definition of a logarithm is $$\log_a b=c \iff a^c=b$$
In your case, this means $$\log_2 \cos x =-\frac12 \iff\cos x=\frac{1}{\sqrt{2}}$$
Do you see how to proceed?